Amrit amrit92

## Proof of Reservoir Sampling
Say we want to generate a set of s elements and that we have already seen n>s elements.

Let's assume that our current s elements have already each been chosen with probability s/n.

By the definition of the algorithm, we choose element n+1 with probability s/(n+1).

Each element already part of our result set has a probability 1/s of being replaced.

The probability that an element from the n-seen result set is replaced in the n+1-seen result set is therefore (1/s)*s/(n+1)=1/(n+1). Conversely, the probability that an element is not replaced is 1-1/(n+1)=n/(n+1).
	Say we want to generate a set of s elements and that we have already seen n>s elements.

	Let's assume that our current s elements have already each been chosen with probability s/n.

	By the definition of the algorithm, we choose element n+1 with probability s/(n+1).

	Each element already part of our result set has a probability 1/s of being replaced.

	The probability that an element from the n-seen result set is replaced in the n+1-seen result set is therefore (1/s)*s/(n+1)=1/(n+1). Conversely, the probability that an element is not replaced is 1-1/(n+1)=n/(n+1).